# From a point on a given circle, tangents are drawn to the ellipse. Need to find locus of chord of contact.

From a point $$O$$ on the circle $$x^2+y^2=d^2$$, tangents $$OP$$ and $$OQ$$ are drawn to the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$, $$a>b$$. Show that the locus of the midpoint of chord PQ is given by $$x^2+y^2=d^2\bigg[\frac{x^2}{a^2}+\frac{y^2}{b^2}\bigg]^2$$

I recognize that the locus of a chord whose midpoint is at $$(h,k)$$ is given by $$\frac{xh}{a^2}+\frac{yk}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$$

I also recognize that PQ is the chord of contact, but to find its equation using the chord of contact formula I would require the coordinates of point O which I do not have.

Here I am getting the equation in terms of $$x,y,h,k$$, but to find the locus I need the equation entirely in the form of $$h,k$$, right? So how do I eliminate $$x,y$$ from the equation of the locus of the midpoint?

This may not be the solution you are looking for:

Let $$\mathcal C=\{x^2+y^2=1\}$$ be the unit circle. Let $$O' = (\alpha, \beta)$$ be any point outside of this circle. Let $$O'P'$$ and $$O'Q'$$ be two tangent line to $$\mathcal C$$. One can check that the midpoint $$m' = (x, y)$$ of $$P'Q'$$ is given by (why?)

$$(x, y) = m' = \frac{1}{\alpha^2+ \beta^2} (\alpha, \beta).$$

Now assume that $$O = (\alpha, \beta)$$ is on the ellipse $$\{ (ax)^2 + (by)^2 = d^2\}$$. Thus $$(a\alpha)^2 + (b\beta)^2 = d^2$$. Then

$$d^2(x^2 + y^2)^2 = \frac{d^2}{(\alpha^2 + \beta^2)^2}$$

and

$$(ax)^2 +(by)^2 = \frac{(a\alpha)^2 + (by)^2}{(\alpha^2 + \beta^2)^2}=\frac{d^2}{(\alpha^2 + \beta^2)^2}$$

Thus the locus of the midpoint $$m'$$ is given by

$$\tag{1} (ax)^2 + (by)^2 = d^2 (x^2+ y^2)^2.$$

The above is related to your question in the following way: Consider the transformation: $$(x, y) \mapsto (x/a, y/b).$$ Under this transformation, the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$ is sent to the unit circle $$\mathcal C$$, while the circle $$x^2 + y^2 = d^2$$ is sent to the ellipse $$(ax)^2 + (by)^2 = d^2$$. The crucial observation is that tangent lines $$OP, OQ$$ are also sent to tangent lines $$O'P', O'Q'$$, and the midpoint $$m$$ of $$PQ$$ are sent to the midpoint $$m'$$ of $$P'Q'$$ (see here). Thus if you take the inverse transformation

$$(x, y) \mapsto (ax, by)$$

Then the locus of $$m'$$ will be sent to the locus of $$m$$. This implies your equation: if you change $$x$$, $$y$$ to $$x/a$$, $$y/b$$ respective in (1), you get

$$x^2 + y^2 = d^2 \left(\frac{x^2}{a^2}+ \frac{y^2}{b^2}\right)^2.$$

• Thanks but I was looking for something starting with what I used Nov 8 '19 at 8:56

Still not what you want, but uses Joachimsthals notations.

The locus is the midpoint of the two intersection points of $$s=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$$ and (from $$s_1^2=s \cdot s_{11}$$) $$(\frac{x(O)x}{a^2}+\frac{y(O)y}{b^2}-1)^2=(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1)(\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}-1)$$ which (since the square roots cancel in $$\frac{x_1+x_2}{2}$$ and $$\frac{y_1+y_2}{2}$$) is $$(x,y)=(\frac{x(O)}{\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}},\frac{y(O)}{\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}}),$$ where $$x(O)^2+y(O)^2=d^2$$ since $$O$$ is on that circle.

Writing $$h=x(O), k=y(O)$$ into M2

R=QQ[a,b,d]
S=R[h,k,x,y,MonomialOrder=>Eliminate 2]
I=ideal(h^2+k^2-d^2,(b^2*h^2+a^2*k^2)*x-a^2*b^2*h,(b^2*h^2+a^2*k^2)*y-a^2*b^2*k)
gens gb I


yields $$a^6b^6d^2(d^2(\frac{x^2}{a^2}+\frac{y^2}{b^2})^2-(x^2+y^2)).$$